z+zi=2+4i. Everything times -i gives -zi+z=-2i+4 or z-zi=4-2i. Done.
@fsponj
2 ай бұрын
Great video as always! I hope your voice returns
@aplusbi
2 ай бұрын
Thank you for the kind words! 🥰
@benpotito
2 ай бұрын
Multiply by -i to equate the second equation to the first, done
@aplusbi
2 ай бұрын
Good thinking
@Qermaq
2 ай бұрын
Math with quirky music makes me remember "as fun tends to infinity"'s channel. Of course we hope you are speaking again soon!
@aplusbi
2 ай бұрын
Thank you! 🥰
@Qermaq
2 ай бұрын
I used the first method, Method 2. Took barely any time at all on a piece of scrap paper. I like problems that you can solve on an index card worth of space. :)
@aplusbi
2 ай бұрын
Excellent!
@mcwulf25
2 ай бұрын
Helps to be familiar with your voice so we can imagine it.
@aplusbi
2 ай бұрын
☺️
@chandraw4262
2 ай бұрын
gws sir ❤❤
@barberickarc3460
2 ай бұрын
I checked if the two functions of Z are linearily dependant with the wronskian and saw they were so you just need to multiply the first one by some constant to get the second. So (z+zi) *A = z-zi Simplify and get A=-i so multiply the first equation to get z-zi=4-2i
@iabervon
2 ай бұрын
I did: z-iz=z(1-i)=z((1-i)/(1+i))(1+i). Now (1-i)/(1+i)=-i (which I visualized in polar form, the moduli cancel out and the arguments subtract), and we have z(1+i), so just multiply that value by -i.
@scottleung9587
2 ай бұрын
I used the second method.
@Alexey_Alex1
2 ай бұрын
Neşeli müzik. Şimdi her zaman böyle olacak mı ?
@aplusbi
2 ай бұрын
Biraz mecbur kaldim ses gittigi icin ama arada olabilir neden olmasin?
@legentlemandriver1034
2 ай бұрын
Four minutes for a simple multiplication: -i(z+zi)=z-zi=4-2i
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